SOLVING LINEAR SYSTEMS BY ADDING OR SUBTRACTING WORKSHEETS

About "Solving linear systems by adding or subtracting worksheets"

Worksheets given in this section would be much useful to the students who would like to practice solving linear equations in two variables.

Solving linear systems by adding or subtracting worksheets - Problems

1. Solve the system of equations by adding. Check your the solution by graphing.

2x - 3y = 12

x + 3y = 6

2. Solve the system of equations by subtracting. Check the solution by graphing.

3x + 3y = 6

3x - y = - 6

3. Sum of the cost price of two products is $50. Sum of the selling price of the same two products is $52. If one is sold at 20% profit and other one is sold at 20% loss, find the cost price of each product.

Solving linear systems by adding or subtracting worksheets - Problems

Problem 1 :

Solve the system of equations by adding. Check your the solution by graphing.

2x - 3y = 12

x + 3y = 6

Answer :

Step 1 :

In the given two equations, the variable y is having the same coefficient (3). And also, the variable y is having different signs.

So we can eliminate the variable y by adding the two equations.

Step 2 :

Solver the resulting equation for the variable x.

3x = 18

Divide both sides by 3.

3x / 3 = 18 / 3

x = 6

Step 3 :

Substitute the value of x into one of the equations to find the value of y.

Solve the system of equations by subtracting. Check the solution by graphing.

3x + 3y = 6

3x - y = - 6

Solution :

Step 1 :

In the given two equations, the variable x is having the same coefficient (3), And also, the variable x is having the same sign in both the equations.

So we can eliminate the variable x by subtracting the two equations.

(3x + 3y) - (3x - y) = (6) - (-6)

3x + 3y - 3x + y = 6 + 6

Simplify.

4y = 12

Divide both sides by 4.

4y / 4 = 12 / 4

y = 3

Step 2 :

Plug y = 3 in one of the equations.

3x - y = - 6

3x - 3 = - 6

Add 3 to both sides.

(3x - 3) + 3 = (-6) + 3

3x - 3 + 3 = -6 + 3

Simplify.

3x = -3

Divide both sides by 3

3x / 3 = -3 / 3

x = - 1

Step 3 :

Write the solution as ordered pair as (x, y).

(-1, 3)

Step 4 :

Check the solution by graphing.

To graph the equations, write them in slope-intercept form.

That is,

y = mx + b

3x + 3y = 6

y = - x + 2

Slope = - 1

y-intercept = 2

3x - y = - 6

y = 3x + 6

Slope = 3

y-intercept = 6

The point of intersection is (-1, 3).

Problem 3 :

Sum of the cost price of two products is $50. Sum of the selling price of the same two products is $52. If one is sold at 20% profit and other one is sold at 20% loss, find the cost price of each product.

Solution :

Step 1 :

Let "x" and "y" be the cost prices of two products.

Then, x + y = 50 --------(1)

Step 2 :

Let us assume that "x" is sold at 20% profit

Then, the selling price of "x" is 120% of "x"

Selling price of "x" = 1.2x

Let us assume that "y" is sold at 20% loss

Then, the selling price of "y" is 80% of "y"

Selling price of "x" = 0.8y

Given : Selling price of "x" + Selling price of "y" = 52

1.2x + 0.8y = 52

To avoid decimal, multiply both sides by 10

12x + 8y = 520

Divide both sides by 4.

3x + 2y = 130 --------(2)

Step 3 :

Eliminate one of the variables to get the value of the other variable.

In (1) and (2), both the variables "x" and "y" are not having the same coefficient.

One of the variables must have the same coefficient.

So multiply both sides of (1) by 2 to make the coefficients of "y" same in both the equations.

(1) ⋅ 2 --------> 2x + 2y = 100 ----------(3)

Variable "y" is having the same sign in both (2) and (3).

To change the sign of "y" in (3), multiply both sides of (3) by negative sign.

- (2x + 2y) = - 100

- 2x - 2y = - 100 --------(4)

Step 4 :

Now, eliminate the variable "y"in (2) and (4) as given below and find the value of "x".